anonymous
  • anonymous
Rationalize the denominator: cubic root of 2y^4/6x^4
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\sqrt[3]{2y ^{4}/6x ^{4}}\]
anonymous
  • anonymous
My instinct is to go down instead of making it \[x ^{64}\] because that would look dumb, unless it's correct but I've never had to decrease... do I just multiply by a negative exponent?
anonymous
  • anonymous
=cube root (2y^4)/cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/cube root(6x^4)*cube root(6x^4) =cube root(2y^4)*cuberoot(6x^4)/6x^4 basically separate top and bottom roots multiply by conjugates fix ugly maths

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jhonyy9
  • jhonyy9
cuberoot(y/3x)^4 =(y/3x)cuberoot(y/3x)=(ycuberoot3xy)/3x
jhonyy9
  • jhonyy9
2/6 simplifie by 2 and get 1/3
jhonyy9
  • jhonyy9
ok ?
anonymous
  • anonymous
Hmm, well that is handy, didn't know oyu could simplify it but what would the correct conjugate be?
anonymous
  • anonymous
you*
jhonyy9
  • jhonyy9
good luck bye
jhonyy9
  • jhonyy9
1/sqrt3 =(sqrt3)/3 ---like example
jhonyy9
  • jhonyy9
ok ?
anonymous
  • anonymous
No actually, I'm confused... can you just tell me what I need to multiply by to get rid of the radical in the denominator?
jhonyy9
  • jhonyy9
1/sqrt3 so when there is radical in denominator multiplie numerator and denominator by conjugate so in this case by sqrt3 and in this way you get in denominator 3 and in numerator sqrt3
jhonyy9
  • jhonyy9
because sqrt3 * sqrt3 =3
anonymous
  • anonymous
I know... but what would the conjugate be. I know how to do all this already i get it I just need to know if the conjugate I am using is correct...
jhonyy9
  • jhonyy9
your answer not is right
jhonyy9
  • jhonyy9
check my answer step by step
razor99
  • razor99
we have to simplify that.
anonymous
  • anonymous
Ok look, the denominator is cubic root of 6x^4 .... so what do I multiply it by to take a perfect cubic root?
anonymous
  • anonymous
sorry that is wrong
anonymous
  • anonymous
\[\sqrt[3]{36x^2}\]
saifoo.khan
  • saifoo.khan
If sat is here, then there was no need of calling me! :)
anonymous
  • anonymous
then you will get in the denominator a perfect cube, namely \[\sqrt[3]{216x^6}\]
anonymous
  • anonymous
He just got here thanks anyway
anonymous
  • anonymous
you got this?
anonymous
  • anonymous
you have \[\sqrt[3]{6x^4}\] so to make it a perfect cube, you need the exponent on the x to be 6, and you have to multiply by 6^2=36 to make the constnant a cube
anonymous
  • anonymous
Wait why does the exponent for x have to be 6?
anonymous
  • anonymous
because you have a cube root so you need a number divisible by 3
anonymous
  • anonymous
i mean an exponent divisible by 3
anonymous
  • anonymous
\[\sqrt[3]{216x^6}=6x^2\]
anonymous
  • anonymous
So that would come out as \[6x ^{2}\] ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ah thank you that is ALL I needed, I guess nobody understood what I meant. Thank you very much
anonymous
  • anonymous
yw

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